Presentation on theme: "Magnetism Chapter 27 opener. Magnets produce magnetic fields, but so do electric currents. An electric current flowing in this straight wire produces a."— Presentation transcript:
1MagnetismChapter 27 opener. Magnets produce magnetic fields, but so do electric currents. An electric current flowing in this straight wire produces a magnetic field which causes the tiny pieces of iron (iron “filings”) to align in the field. We shall see in this Chapter how magnetic field is defined, and that the magnetic field direction is along the iron filings. The magnetic field lines due to the electric current in this long wire are in the shape of circles around the wire. We also discuss how magnetic fields exert forces on electric currents and on charged particles, as well as useful applications of the interaction between magnetic fields and electric currents and moving electric charges.
2Magnets and Magnetic Fields Electric Currents Produce Magnetic FieldsForce on an Electric Current in a Magnetic Field; Definition of BForce on an Electric Charge Moving in a Magnetic FieldDiscovery and Properties of the ElectronMass Spectrometer
3Magnets and Magnetic Fields Magnets have two ends – poles – called north and south.Like poles repel; unlike poles attract.Figure Like poles of a magnet repel; unlike poles attract. Red arrows indicate force direction.
4Magnets and Magnetic Fields However, if you cut a magnet in half, you don’t get a north pole and a south pole – you get two smaller magnets.Figure If you split a magnet, you won’t get isolated north and south poles; instead, two new magnets are produced, each with a north and a south pole.
5Magnets and Magnetic Fields Magnetic fields can be visualized using magnetic field lines, which are always closed loops.Figure (a) Visualizing magnetic field lines around a bar magnet, using iron filings and compass needles. The red end of the bar magnet is its north pole. The N pole of a nearby compass needle points away from the north pole of the magnet. (b) Magnetic field lines for a bar magnet.
6Magnets and Magnetic Fields Magnetic field lines:Start from the north pole and end at the south pole.Do not intersectTangent to the linesDensity of the lines
7Magnets and Magnetic Fields The Earth’s magnetic field is similar to that of a bar magnet.Note that the Earth’s “North Pole” is really a south magnetic pole, as the north ends of magnets are attracted to it.Figure The Earth acts like a huge magnet; but its magnetic poles are not at the geographic poles, which are on the Earth’s rotation axis.
8Magnets and Magnetic Fields A uniform magnetic field is constant in magnitude and direction.The field between these two wide poles is nearly uniform.Figure Magnetic field between two wide poles of a magnet is nearly uniform, except near the edges.
9Electric Currents Produce Magnetic Fields Experiment shows that an electric current produces a magnetic field. The direction of the field is given by a right-hand rule.Figure (a) Deflection of compass needles near a current-carrying wire, showing the presence and direction of the magnetic field. (b) Magnetic field lines around an electric current in a straight wire. (c) Right-hand rule for remembering the direction of the magnetic field: when the thumb points in the direction of the conventional current, the fingers wrapped around the wire point in the direction of the magnetic field. See also the Chapter-Opening photo.
10Electric Currents Produce Magnetic Fields Here we see the field due to a current loop; the direction is again given by a right-hand rule.Figure Magnetic field lines due to a circular loop of wire.Figure Right-hand rule for determining the direction of the magnetic field relative to the current.
11Definition of the Magnetic Field A magnet exerts a force on a current-carrying wire. The direction of the force is given by a right-hand rule.Figure (a) Force on a current-carrying wire placed in a magnetic field B; (b) same, but current reversed; (c) right-hand rule for setup in (b).
12Definition of the Magnetic Field The force on the wire depends on the current, the length of the wire, the magnetic field, and its orientation:This equation defines the magnetic field B.In vector notation:
13Definition of the Magnetic Field Unit of B: the tesla, T:1 T = 1 N/A·m.Another unit sometimes used: the gauss (G):1 G = 10-4 T.
14Force on an Electric Current Magnetic Force on a current-carrying wire.A wire carrying a 30-A current has a length l = 12 cm between the pole faces of a magnet at an angle θ = 60°, as shown. The magnetic field is approximately uniform at T. We ignore the field beyond the pole pieces. What is the magnitude of the force on the wire?Solution: F = IlBsin θ = 2.8 N.
15Force on an Electric Current Measuring a magnetic field.A rectangular loop of wire hangs vertically as shown. A magnetic field B is directed horizontally, perpendicular to the wire, and points out of the page at all points. The magnetic field is very nearly uniform along the horizontal portion of wire ab (length l = 10.0 cm) which is near the center of the gap of a large magnet producing the field. The top portion of the wire loop is free of the field. The loop hangs from a balance which measures a downward magnetic force (in addition to the gravitational force) of F = 3.48 x 10-2 N when the wire carries a current I = A. What is the magnitude of the magnetic field B?Solution: The wire is perpendicular to the field (the vertical wires feel no force), so B = F/Il = 1.42 T.
16Force on an Electric Current Magnetic Force on a semicircular wire.A rigid wire, carrying a current I, consists of a semicircle of radius R and two straight portions as shown. The wire lies in a plane perpendicular to a uniform magnetic field B0. Note choice of x and y axis. The straight portions each have length l within the field. Determine the net force on the wire due to the magnetic field B0.Solution: The forces on the straight sections are equal and opposite, and cancel. The force dF on a segment dl of the wire is IB0R dφ and is perpendicular to the plane of the diagram. Therefore, F = ∫dF = IB0R∫sin φ dφ = 2IB0R.
17Force on an Electric Charge The force on a moving charge is related to the force on a current:Once again, the direction is given by a right-hand rule.Figure Force on charged particles due to a magnetic field is perpendicular to the magnetic field direction.
18Force on an Electric Charge Negative charge near a magnet.A negative charge -Q is placed at rest near a magnet. Will the charge begin to move? Will it feel a force? What if the charge were positive, +Q?Solution: There is no force on a motionless charge, be it positive or negative, so in neither case will it begin to move.
19Force on an Electric Charge Magnetic force on a proton.A magnetic field exerts a force of 8.0 x N toward the west on a proton moving vertically upward at a speed of 5.0 x 106 m/s (a). When moving horizontally in a northerly direction, the force on the proton is zero (b). Determine the magnitude and direction of the magnetic field in this region. (The charge on a proton is q = +e = 1.6 x C.)Solution: Since the force is zero when the proton is moving north, the field must point in the north-south direction. In order for the force to be to the west when the proton is moving up, the field must point north. B = F/qv = 0.10 T.
20Force on an Electric Charge Magnetic force on ions during a nerve pulse.Estimate the magnetic force due to the Earth’s magnetic field on ions crossing a cell membrane during an action potential. Assume the speed of the ions is 10-2 m/s.Solution: F = qvB; assume B = 10-4 T, and use the electron charge for q. F = N.
21Force on an Electric Charge If a charged particle is moving perpendicular to a uniform magnetic field, its path will be a circle.Figure Force exerted by a uniform magnetic field on a moving charged particle (in this case, an electron) produces a circular path.
22Force on an Electric Charge Electron’s path in a uniform magnetic field.An electron travels at 2.0 x 107 m/s in a plane perpendicular to a uniform T magnetic field. Describe its path quantitatively.Solution: The magnetic force keeps the particle moving in a circle, so mv2/r = qvB. Solving for r gives r = mv/qB = 1.1 cm.
23Force on an Electric Charge Stopping charged particles.Can a magnetic field be used to stop a single charged particle, as an electric field can?Solution: No, because the force is always perpendicular to the velocity. In fact, a uniform magnetic field cannot change the speed of a charged particle, only its direction.
25Force on an Electric Charge A helical path.What is the path of a charged particle in a uniform magnetic field if its velocity is not perpendicular to the magnetic field?Solution: The path is a helix – the component of velocity parallel to the magnetic field does not change, and the velocity in the plane perpendicular to the field describes a circle.
26Force on an Electric Charge The aurora borealis (northern lights) is caused by charged particles from the solar wind spiraling along the Earth’s magnetic field, and colliding with air molecules.Figure (a) Diagram showing a negatively charged particle that approaches the Earth and is “captured” by the magnetic field of the Earth. Such particles follow the field lines toward the poles as shown. (b) Photo of aurora borealis (here, in Kansas, a rare sight).
27Force on an Electric Charge Velocity selector, or filter: crossed E and B fields.Some electronic devices and experiments need a beam of charged particles all moving at nearly the same velocity. This can be achieved using both a uniform electric field and a uniform magnetic field, arranged so they are at right angles to each other. Particles of charge q pass through slit S1 and enter the region where E points into the page and B points down from the positive plate toward the negative plate. If the particles enter with different velocities, show how this device “selects” a particular velocity, and determine what this velocity is.Figure 27-21: A velocity selector: if v = E/B, the particles passing through S1 make it through S2.Solution: Only the particles whose velocities are such that the magnetic and electric forces exactly cancel will pass through both slits. We want qE = qvB, so v = E/B.
28ElectronElectrons were first observed in cathode ray tubes. These tubes had a very small amount of gas inside, and when a high voltage was applied to the cathode, some “cathode rays” appeared to travel from the cathode to the anode.Figure Discharge tube. In some models, one of the screens is the anode (positive plate).
29ElectronThe value of e/m for the cathode rays was measured in 1897 using the apparatus below; it was then that the rays began to be called electrons.Figure goes here.Figure Cathode rays deflected by electric and magnetic fields.
31ElectronMillikan measured the electron charge directly shortly thereafter, using the oil-drop apparatus diagrammed below, and showed that the electron was a constituent of the atom (and not an atom itself, as its mass is far too small).The currently accepted values of the electron mass and charge arem = 9.1 x kge = 1.6 x CFigure Millikan’s oil-drop experiment.
32Mass SpectrometerA mass spectrometer measures the masses of atoms. If a charged particle is moving through perpendicular electric and magnetic fields, there is a particular speed at which it will not be deflected, which then allows the measurement of its mass:
33Mass SpectrometerAll the atoms reaching the second magnetic field will have the same speed; their radius of curvature will depend on their mass.Figure Bainbridge-type mass spectrometer. The magnetic fields B and B’ point out of the paper (indicated by the dots), for positive ions.
35Mass Spectrometer Mass spectrometry. Carbon atoms of atomic mass 12.0 u are found to be mixed with another, unknown, element. In a mass spectrometer with fixed B′, the carbon traverses a path of radius 22.4 cm and the unknown’s path has a 26.2-cm radius. What is the unknown element? Assume the ions of both elements have the same charge.Solution: The ratio of masses is equal to the ratio of the radii, or Therefore the unknown element has a mass of 14.0 u, and is probably nitrogen.
36Summary Magnets have north and south poles. Like poles repel, unlike attract.Unit of magnetic field: tesla.Electric currents produce magnetic fields.A magnetic field exerts a force on an electric current:
37SummaryA magnetic field exerts a force on a moving charge:
38Magnetic FieldChapter 28 opener. A long coil of wire with many closely spaced loops is called a solenoid. When a long solenoid carries an electric current, a nearly uniform magnetic field is produced within the loops as suggested by the alignment of the iron filings in this photo. The magnitude of the field inside a solenoid is readily found using Ampère’s law, one of the great general laws of electromagnetism, relating magnetic fields and electric currents. We examine these connections in detail in this Chapter, as well as other means for producing magnetic fields.
39Magnetic Field Due to a Straight Wire Force between Two Parallel WiresDefinitions of the Ampere and the CoulombAmpère’s LawMagnetic Field of a Solenoid and a ToroidElectromagnets and Solenoids – Applications
40Field Due to a Straight Wire The magnetic field due to a straight wire is inversely proportional to the distance from the wire:The constant μ0 is called the permeability of free space, and has the valueμ0 = 4π x 10-7 T·m/A.Figure Same as Fig. 27–8b. Magnetic field lines around a long straight wire carrying an electric current I.
41Field Due to a Straight Wire Calculation of B near a wire.An electric wire in the wall of a building carries a dc current of 25 A vertically upward. What is the magnetic field due to this current at a point P 10 cm due north of the wire?Solution: B = μ0I/2πr = 5.0 x 10-5 T
42Field Due to a Straight Wire Magnetic field midway between two currents.Two parallel straight wires 10.0 cm apart carry currents in opposite directions. Current I1 = 5.0 A is out of the page, and I2 = 7.0 A is into the page. Determine the magnitude and direction of the magnetic field halfway between the two wires.Figure Example 28–2. Wire 1 carrying current I1 out towards us, and wire 2 carrying current I2 into the page, produce magnetic fields whose lines are circles around their respective wires.Solution: As the figure shows, the two fields are in the same direction midway between the wires. Therefore, the total field is the sum of the two, and points upward: B1 = 2.0 x 10-5 T; B2 = 2.8 x 10-5 T; so B = 4.8 x 10-5 T.
43Field Due to a Straight Wire Magnetic field due to four wires.This figure shows four long parallel wires which carry equal currents into or out of the page. In which configuration, (a) or (b), is the magnetic field greater at the center of the square?Solution: The fields cancel in (b) but not in (a); therefore the field is greater in (a).
44Force between Parallel Wires The magnetic field produced at the position of wire 2 due to the current in wire 1 isThe force this field exerts on a length l2 of wire 2 isFigure (a) Two parallel conductors carrying currents I1 and I2. (b) Magnetic field B1 produced by I1 (Field produced by I2 is not shown.) B1 points into page at position of I2.
45Force between Parallel Wires Parallel currents attract; antiparallel currents repel.Figure (a) Parallel currents in the same direction exert an attractive force on each other. (b) Antiparallel currents (in opposite directions) exert a repulsive force on each other.
46Force between Parallel Wires Force between two current-carrying wires.The two wires of a 2.0-m-long appliance cord are 3.0 mm apart and carry a current of 8.0 A dc. Calculate the force one wire exerts on the other.Solution: F = 8.5 x 10-3 N, and is repulsive. Appliances powered by 12V dc current are marketed to truckers, campers, people running on solar power (especially off-grid), among others.
47Force between Parallel Wires Suspending a wire with a current.A horizontal wire carries a current I1 = 80 A dc. A second parallel wire 20 cm below it must carry how much current I2 so that it doesn’t fall due to gravity? The lower wire has a mass of 0.12 g per meter of length.Solution: The magnetic force due to the current in the wire must be equal and opposite to the gravitational force on the second wire. We need a definite length of wire, as the forces on an infinitely long wire will be infinite; choose 1 meter for simplicity. Substitution gives I2 = 15 A.
48Definitions of the Ampere and the Coulomb The ampere is officially defined in terms of the force between two current-carrying wires:One ampere is defined as that current flowing in each of two long parallel wires 1 m apart, which results in a force of exactly 2 x 10-7 N per meter of length of each wire.The coulomb is then defined as exactly one ampere-second.
49Biot-Savart LawThe Biot-Savart law gives the magnetic field due to an infinitesimal length of current; the total field can then be found by integrating over the total length of all currents:Figure Biot-Savart law: the field at P due to current element Idl is dB = (μ0I/4π)(dl x r/r2)
50Biot-Savart Law B due to current I in straight wire. For the field near a long straight wire carrying a current I, show that the Biot-Savart law gives B = μ0I/2πr.Figure Determining B due to a long straight wire using the Biot-Savart law.Solution: Starting with the initial equation for the magnitude of B (integrating over y from -∞ to ∞), substitute y = -R/tan θ so dy = r2 dθ/R. Integrating over θ from 0 to π then gives the result.
52Biot-Savart Law Current loop. Determine B for points on the axis of a circular loop of wire of radius R carrying a current I.Solution: (see figure) The perpendicular components of B cancel, leaving only the component along the axis. dB = μ0I dl/4πr2; the component we want is dB cos φ. Since all segments of the current are the same distance from the point P, each contributes the same amount to B, and the integral is only over dl, giving 2πR.
53Solution:The perpendicular components of B cancel, leaving along only the component the axis.
54Biot-Savart Law B due to a wire segment. One quarter of a circular loop of wire carries a current I. The current I enters and leaves on straight segments of wire, as shown; the straight wires are along the radial direction from the center C of the circular portion. Find the magnetic field at point C.Solution: The straight segments of wire produce no field at C, as the current is parallel to r. On the curved segment, all points are the same distance (R) from C; the integral is therefore only over dl, and the result is B = μ0I/8R.
55Solution:The straight segments of wire produce no field at C, as the current is parallel to r.
56Ampère’s LawAmpère’s law relates the magnetic field around a closed loop to the total current flowing through the loop:Figure Arbitrary path enclosing a current, for Ampère’s law. The path is broken down into segments of equal length Δl.This integral is taken around the edge of the closed loop.
57Ampère’s LawUsing Ampère’s law to find the field around a long straight wire:Use a circular path with the wire at the center; then B is tangent to dl at every point. The integral then givesFigure Circular path of radius r.so B = μ0I/2πr, as before.
58Ampère’s Law Field inside and outside a wire. A long straight cylindrical wire conductor of radius R carries a current I of uniform current density in the conductor. Determine the magnetic field due to this current at (a) points outside the conductor (r > R) and (b) points inside the conductor (r < R). Assume that r, the radial distance from the axis, is much less than the length of the wire. (c) If R = 2.0 mm and I = 60 A, what is B at r = 1.0 mm, r = 2.0 mm, and r = 3.0 mm?Solution: We choose a circular path around the wire; if the wire is very long the field will be tangent to the path.a. The enclosed current is the total current; this is the same as a thin wire. B = μ0I/2πr.b. Now only a fraction of the current is enclosed within the path; if the current density is uniform the fraction of the current enclosed is the fraction of area enclosed: Iencl = Ir2/R2. Substituting and integrating gives B = μ0Ir/2πR2.c. 1 mm is inside the wire and 3 mm is outside; 2 mm is at the surface (so the two results should be the same). Substitution gives B = 3.0 x 10-3 T at 1.0 mm, 6.0 x 10-3 T at 2.0 mm, and 4.0 x 10-3 T at 3.0 mm.
59Ampère’s Law Coaxial cable. A coaxial cable is a single wire surrounded by a cylindrical metallic braid. The two conductors are separated by an insulator. The central wire carries current to the other end of the cable, and the outer braid carries the return current and is usually considered ground. Describe the magnetic field (a) in the space between the conductors, and (b) outside the cable.Solution: a. Between the conductors, the field is solely due to the inner conductor, and is that of a long straight wire.b. Outside the cable the field is zero.
60Ampère’s Law A nice use for Ampère’s law. Use Ampère’s law to show that in any region of space where there are no currents the magnetic field cannot be both unidirectional and nonuniform as shown in the figure.Solution: Use Ampère’s law to integrate around the path shown. The two vertical segments contribute zero to the integral, as the path is perpendicular to the field. The contributions from the horizontal segments would cancel if the field were uniform; if it is not, they don’t. This is inconsistent with Ampère’s law, as there is no current enclosed by the path.
61Ampère’s Law Solving problems using Ampère’s law: Ampère’s law is only useful for solving problems when there is a great deal of symmetry. Identify the symmetry.Choose an integration path that reflects the symmetry (typically, the path is along lines where the field is constant and perpendicular to the field where it is changing).Use the symmetry to determine the direction of the field.Determine the enclosed current.
62Magnetic Field of a Solenoid and a Toroid A solenoid is a coil of wire containing many loops. To find the field inside, we use Ampère’s law along the path indicated in the figure.Figure 28-16: Cross-sectional view into a solenoid. The magnetic field inside is straight except at the ends. Red dashed lines indicate the path chosen for use in Ampère’s law.
63Solenoid and ToroidThe field is zero outside the solenoid, and the path integral is zero along the vertical lines, so the field is (n is the number of loops per unit length)
64Solenoid and Toroid Field inside a solenoid. A thin 10-cm-long solenoid used for fast electromechanical switching has a total of 400 turns of wire and carries a current of 2.0 A. Calculate the field inside near the center.Solution: Substitution gives B = 1.0 x 10-2 T.
65Solenoid and ToroidToroid. Use Ampère’s law to determine the magnetic field (a) inside and (b) outside a toroid, which is like a solenoid bent into the shape of a circle as shown.Figure (a) A toroid. (b) A section of the toroid showing direction of the current for three loops.Solution: a. Use path 1 in the figure. The enclosed current is NI (only the inside part of each loop is enclosed, and all the currents run in the same direction). This gives B = μ0NI/2πr.
66Electromagnets and Solenoids – Applications Remember that a solenoid is a long coil of wire. If it is tightly wrapped, the magnetic field in its interior is almost uniform.Figure Magnetic field of a solenoid. The north pole of this solenoid, thought of as a magnet, is on the right, and the south pole is on the left.
67Electromagnets and Solenoids – Applications If a piece of iron is inserted in the solenoid, the magnetic field greatly increases. Such electromagnets have many practical applications.Figure Solenoid used as a doorbell.
68SummaryMagnitude of the field of a long, straight current-carrying wire:The force of one current-carrying wire on another defines the ampere.Ampère’s law:
69Summary Magnetic field inside a solenoid: Biot-Savart law: Ferromagnetic materials can be made into strong permanent magnets.